Hello all, Recently I was attempting the following problem: Given an array of integers, arr. Find sum of floor of (arr[i]/arr[j]) for all pairs of indices (i,j).

e.g. arr[]={1,2,3,4,5}, Sum=27.

I could only think of naive O(n^2) solution. Is there any other better approach?

Thanks in advance.

Is there any constraint on the values that the elements of the array may have ?

Okay, the size of array n is 1<=n<=1e5 and each element arr[i] is 1<=arr[i]<=1e5.

Firstly, precalculate b[i] values which is equal to count of numbers ≥

i, it can be done inO(MX). Then for each number, go like a sieve and now calculating how answer change is easy since we know the count of elements between this and next multiple.Code with a little commentNote: I assume there are no same elements in the array. Otherwise, the code can easily be modified.

You're using suffix sums. I think use of prefix sums would have been more intuitive. But anyways, good logic.

Great solution!! Thanks for the help

I still dont get it. can you give explanation with an example?

Consider a number i so floor of all numbers between i and 2*i(2*i not included) is gonna to to contribute 1 to the sum.And all the numbers between 2*i and 3*i(3*i not included) is gonna to contribute 2 to the sum and so on. The mul var is taking in to account this thing.

I think this solution is wrong because if the array is [1,2,3,4,5] then the output should be 22 but the output of your code is 27.

The correct output is $$$27$$$. Perhaps you are not considering the case where $$$i = j$$$?

Link to the problem

Solution

Good solution but value of n is too much higher instead you can set value of n in this way and u can see the running time difference 9 sec to 6 sec for all test cases.

You can use another approach. 1. Sort elements. 2. For elements less than sqrt(n) you can run through the array each time(Not forget to remember result of each run, because elements may be repeated). 3. For elements greater than sqrt(n), let's say x, you may use binary search to find the number of elements between xi and x(i+1) (say res), and add res*i to the answer. So overall time complexity will be no less than linearithmic and no greater than O(n*sqrt(n)*log(n)).